De Moivre's Theorem MCQ Quiz in తెలుగు - Objective Question with Answer for De Moivre's Theorem - ముఫ్త్ [PDF] డౌన్‌లోడ్ కరెన్

Last updated on Mar 15, 2025

పొందండి De Moivre's Theorem సమాధానాలు మరియు వివరణాత్మక పరిష్కారాలతో బహుళ ఎంపిక ప్రశ్నలు (MCQ క్విజ్). వీటిని ఉచితంగా డౌన్‌లోడ్ చేసుకోండి De Moivre's Theorem MCQ క్విజ్ Pdf మరియు బ్యాంకింగ్, SSC, రైల్వే, UPSC, స్టేట్ PSC వంటి మీ రాబోయే పరీక్షల కోసం సిద్ధం చేయండి.

Latest De Moivre's Theorem MCQ Objective Questions

Top De Moivre's Theorem MCQ Objective Questions

De Moivre's Theorem Question 1:

If  are respectively the least positive and greatest negative integer values of  such that  then 

  1. 4
  2. 0
  3. 6
  4. 2

Answer (Detailed Solution Below)

Option 1 : 4

De Moivre's Theorem Question 1 Detailed Solution

Calculation

Using De Moivre's Theorem:

Comparing with -i, we have:

This implies:

⇒ 

⇒ 

⇒ 

⇒ 

For the least positive integer value of k, let n = 0:

Let n = 1:

Let n = 2:

So, m = 3.

For the greatest negative integer value of k, we can analyze the values of k for negative n.

For n = -1:

So, n = -1.

Hence option 1 is correct

De Moivre's Theorem Question 2:

For z ∈ , if (1 + z)n = 1 + nc1z + nc2z2 + .... ncnzn and , then k =

  1. 25
  2. 100
  3. 50
  4. 75

Answer (Detailed Solution Below)

Option 3 : 50

De Moivre's Theorem Question 2 Detailed Solution

Concept:

Euler's theorem:

e = cosθ + isinθ 

De Moivre's Theorem:

(cosθ + isinθ)ncos(nθ) + isin(nθ) 

Calculation:

Given, 

Putting z = eir, we get:

...(i)

Now 

, where Img(z) represents the imaginary part of z

 =  (given)

Comparing both sides we get, k = 50

∴ The value of k is 50.

De Moivre's Theorem Question 3:

If z = e and  =  

then  =

  1. 0
  2. 1
  3. 2
  4. 3

Answer (Detailed Solution Below)

Option 2 : 1

De Moivre's Theorem Question 3 Detailed Solution

Concept:

  • If  then by componendo and dividendo rule  

Calculation:

Given z = e and   = 

⇒    = 

We know that if  then 

⇒  = 

⇒  =   [e = cos θ + i sin θ ]

⇒  =  

We know zr = eirθ and in the R.H.S the value of r are 2, 3 and 5

∴ a0 + b0 = a1 + b1 = a4 + b4 = a6 + b6 = a7 + b7 = a8 + b8 = a9 + b9 = a10 + b10 = 0

∴ a2 + b2 = 2 and a3 + b3 = 3 and a5 + b5 = 5 .

⇒  = 1

Required value of the expression is 1

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